Question

1.z=m+a+b

Answer 1

It's literal equations

Answer 2

Ok, do we have a word problem so we can plug the numbers in?

Answer 3

Are there any numbers representing these variables?

Answer 4

no my teacher didn't put numbers on them

Answer 5

So what variable are we solving for then?

Answer 6

a

Answer 7

Ok so since this is a simple addition equation, all we would have to do is just subtract M and B from the right side and put it on the left side. This should isolate A on a side by itself.

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @animegirl0
As written this isn't really rigorous. You hint at using induction to prove this, so you should actually demonstrate (by writing down an induction argument) that it works, instead of just showing it for a few cases.

That is, suppose that
1(1−z)n=1+(nn−1)z+(n+1n−1)z2+(n+2n−1)z3+⋯
and then derive from this that
1(1−z)n+1=1+(n+1n)z+(n+2n)z2+(n+3n)z3+⋯
You should also stipulate that this is only valid when |z|<1.
\(\color{#0cbb34}{\text{End of Quote}}\)
Nice copy and paste. this doesn't even relate to this question.

Answer 9

ok thank you

Answer 10

np. What did you get as ur answer?

Answer 11

I got z-b=a?

Answer 12

What happened to the variable M?

Answer 13

oh z-m-b=a then

Answer 14

That is correct good job